Expanding the <spatial projection tensor> gives
$$
\Theta_\mu
=\perp^\rho{}_\mu Z_\rho
=Z_\mu+n_\mu n^\rho Z_\rho
=Z_\mu-n_\mu\Theta.
$$
Rearrangement yields the normal-plus-spatial decomposition
$$
\boxed{Z_\mu=\Theta_\mu+n_\mu\Theta.}
$$
It also makes $n^\mu\Theta_\mu=0$ immediate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 1 / a / i / Solution

{title2=$\perp^\mu{}_\nu=\delta^\mu{}_\nu+n^\mu n_\nu$}

For a unit timelike normal $n^\mu n_\mu=-1$, the spatial projection tensor annihilates $n^\mu$ and projects every spacetime tensor onto a spatial hypersurface.


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